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Dynamic Programming Interview

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Dynamic Programming Interview in Action

Problem: "0/1 Knapsack Problem - Find maximum value within weight capacity"

Interviewer: "You have a knapsack with capacity W and n items, each with weight w[i] and value v[i]. Find the maximum value you can achieve without exceeding the weight limit. Each item can be used at most once."

Problem Analysis & Approach

Example Input:

  • Capacity: W = 50
  • Items: [(weight=10, value=60), (weight=20, value=100), (weight=30, value=120)]
  • Expected Output: Maximum value = 220 (items 2 and 3)

DP State Definition:

  • dp[i][w] = maximum value using first i items with weight limit w
  • Base case: dp[0][w] = 0 (no items = no value)
  • Recurrence: dp[i][w] = max(exclude item i, include item i)

1. Recursive Solution (Brute Force)

// Recursive approach - explores all possibilities function knapsackRecursive(weights, values, capacity, n) { // Base case: no items or no capacity if (n === 0 || capacity === 0) { return 0; } // If current item weight exceeds capacity, can't include it if (weights[n - 1] > capacity) { return knapsackRecursive(weights, values, capacity, n - 1); } // Choice: include current item or exclude it const includeItem = values[n - 1] + knapsackRecursive(weights, values, capacity - weights[n - 1], n - 1); const excludeItem = knapsackRecursive(weights, values, capacity, n - 1); // Return maximum of both choices return Math.max(includeItem, excludeItem); } // Time Complexity: O(2^n) - exponential // Space Complexity: O(n) - recursion depth

Recursive Solution Analysis:

  • Problem: Exponential time complexity due to overlapping subproblems
  • Example: Computing knapsack(items=3, capacity=10) multiple times
  • Solution: Use memoization to cache results and avoid recomputation
  • Pattern Recognition: Choice at each step + optimal substructure = DP candidate

2. Top-Down DP (Memoization)

// Memoization - cache recursive results function knapsackMemo(weights, values, capacity) { const n = weights.length; // Create memoization table: -1 means not computed const memo = Array(n + 1).fill().map(() => Array(capacity + 1).fill(-1) ); function dp(i, w) { // Base case if (i === 0 || w === 0) { return 0; } // Return cached result if already computed if (memo[i][w] !== -1) { return memo[i][w]; } // Current item exceeds weight limit if (weights[i - 1] > w) { memo[i][w] = dp(i - 1, w); } else { // Choose max of including or excluding current item memo[i][w] = Math.max( dp(i - 1, w), // Exclude item i values[i - 1] + dp(i - 1, w - weights[i - 1]) // Include item i ); } return memo[i][w]; } return dp(n, capacity); } // Time Complexity: O(n * W) - each subproblem computed once // Space Complexity: O(n * W) - memoization table + O(n) recursion

3. Bottom-Up DP (Tabulation)

// Tabulation - iterative bottom-up approach function knapsackTabulation(weights, values, capacity) { const n = weights.length; // Create DP table: dp[i][w] = max value with first i items and weight w const dp = Array(n + 1).fill().map(() => Array(capacity + 1).fill(0) ); // Fill the DP table for (let i = 1; i <= n; i++) { for (let w = 1; w <= capacity; w++) { // Current item index (0-based) const currentWeight = weights[i - 1]; const currentValue = values[i - 1]; if (currentWeight <= w) { // Can include current item - choose max dp[i][w] = Math.max( dp[i - 1][w], // Exclude current item currentValue + dp[i - 1][w - currentWeight] // Include current item ); } else { // Cannot include current item dp[i][w] = dp[i - 1][w]; } } } // Backtrack to find selected items (optional) const selectedItems = []; let i = n, w = capacity; while (i > 0 && w > 0) { // If value came from including current item if (dp[i][w] !== dp[i - 1][w]) { selectedItems.push(i - 1); // 0-based index w -= weights[i - 1]; } i--; } return { maxValue: dp[n][capacity], selectedItems: selectedItems.reverse() }; } // Time Complexity: O(n * W) - two nested loops // Space Complexity: O(n * W) - DP table only

DP Table Visualization

Knapsack DP Table Example (Weights=[1,3,4,5], Values=[1,4,5,7], Capacity=7):

Items:  0   1   2   3   4  (item indices)
       [0] [1] [4] [5] [7] (values)
       [0] [1] [3] [4] [5] (weights)

     w: 0   1   2   3   4   5   6   7
i=0:    0   0   0   0   0   0   0   0   (no items)
i=1:    0   1   1   1   1   1   1   1   (item 0: w=1,v=1)
i=2:    0   1   1   4   5   5   5   5   (item 1: w=3,v=4)
i=3:    0   1   1   4   5   6   6   9   (item 2: w=4,v=5)
i=4:    0   1   1   4   5   7   8   9   (item 3: w=5,v=7)

Final Answer: dp[4][7] = 9
Selected Items: [0, 1] (weights: 1+3=4, values: 1+4=5) 
                or [2] (weight: 4, value: 5) 
                or [3] (weight: 5, value: 7)
Optimal: Item 3 alone gives value 7, or items 0+2 give value 6
Actually: Items 1+2 give value 4+5=9 with weight 3+4=7 ✓

4. Space-Optimized DP

// Space-optimized version using only 1D array function knapsackOptimized(weights, values, capacity) { const n = weights.length; // Only need current row of DP table let dp = Array(capacity + 1).fill(0); // Process each item for (let i = 0; i < n; i++) { const weight = weights[i]; const value = values[i]; // Traverse from right to left to avoid using updated values for (let w = capacity; w >= weight; w--) { dp[w] = Math.max(dp[w], dp[w - weight] + value); } } return dp[capacity]; } // Time Complexity: O(n * W) - same as 2D version // Space Complexity: O(W) - only single array needed!

DP Approach Comparison

Approach Time Complexity Space Complexity Pros & Cons
Recursive O(2^n) O(n) Easy to understand, exponential time
Memoization O(n × W) O(n × W) + O(n) Top-down, natural recursion + cache
Tabulation O(n × W) O(n × W) Bottom-up, can track solution path
Space-Optimized O(n × W) O(W) Minimal memory, harder to track items

Note: n = number of items, W = knapsack capacity

Key DP Interview Insights:

1. Problem Recognition:

  • Optimal Substructure: Optimal solution contains optimal solutions to subproblems
  • Overlapping Subproblems: Same subproblems solved multiple times
  • Choice at each step: Include/exclude item creates decision tree

2. State Definition Strategy:

  • What varies? Item index (i) and remaining capacity (w)
  • What do we want? Maximum value achievable
  • State: dp[i][w] = max value with first i items, capacity w

3. Implementation Tips:

  • Start simple: Write recursive solution first, then optimize
  • Base cases: Handle edge cases (no items, no capacity)
  • Direction: In 1D optimization, iterate backwards to avoid overwriting
  • Backtracking: If solution path needed, keep 2D table

Follow-up Interview Questions:

  • "What if we can use unlimited quantity of each item?" (Unbounded Knapsack)
  • "How would you handle fractional knapsack?" (Greedy approach)
  • "Can you reduce space complexity further?" (Use only two 1D arrays)
  • "How would you handle negative weights or values?"

🧮 Classic DP Problems

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📊 Pattern Recognition

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